Help pls pls pls pls.
Giải chi tiết.
View attachment 171804
Một cách khác là sử dụng công thức giảm bậc (2 lần với bài này):
$${\displaystyle\int}\sec^{\mathtt{n}}\left(x\right)\,\mathrm{d}x=\class{steps-node}{\cssId{steps-node-1}{\dfrac{\mathtt{n}-2}{\mathtt{n}-1}}}{\displaystyle\int}\sec^{\mathtt{n}-2}\left(x\right)\,\mathrm{d}x+\dfrac{\sec^{\mathtt{n}-2}\left(x\right)\tan\left(x\right)}{\mathtt{n}-1}$$
Ta được:
$${\displaystyle\int}\dfrac{1}{\cos^5\left(x\right)}\,\mathrm{d}x = {\displaystyle\int}\sec^5\left(x\right)\,\mathrm{d}x$$
$$=\dfrac{\sec^3\left(x\right)\tan\left(x\right)}{4}+\class{steps-node}{\cssId{steps-node-2}{\dfrac{3}{4}}}{\displaystyle\int}\sec^3\left(x\right)\,\mathrm{d}x$$
$$=\dfrac{\sec^3\left(x\right)\tan\left(x\right)}{4}+\class{steps-node}{\cssId{steps-node-2}{\dfrac{3}{4}}} \left(\dfrac{\sec\left(x\right)\tan\left(x\right)}{2}+\class{steps-node}{\cssId{steps-node-3}{\dfrac{1}{2}}}{\displaystyle\int}\sec\left(x\right)\,\mathrm{d}x\right)$$
Ta có:
$${\displaystyle\int}\sec\left(x\right)\,\mathrm{d}x$$
$$={\displaystyle\int}\dfrac{\sec\left(x\right)\left(\tan\left(x\right)+\sec\left(x\right)\right)}{\tan\left(x\right)+\sec\left(x\right)}\,\mathrm{d}x ={\displaystyle\int}\dfrac{\sec\left(x\right)\tan\left(x\right)+\sec^2\left(x\right)}{\tan\left(x\right)+\sec\left(x\right)}\,\mathrm{d}x \,\, (1)$$
Đặt $u=\tan\left(x\right)+\sec\left(x\right) \implies \mathrm{d}u= \sec\left(x\right)\tan\left(x\right)+\sec^2\left(x\right) \mathrm{d}x$
$(1)$ trở thành:
$${\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u = \ln\left(u\right) = \ln\left(|\tan\left(x\right)+\sec\left(x\right)|\right)$$
Vậy kết quả cuối cùng là:
$$\dfrac{\sec^3\left(x\right)\tan\left(x\right)}{4}+\class{steps-node}{\cssId{steps-node-2}{\dfrac{3}{4}}} \left(\dfrac{\sec\left(x\right)\tan\left(x\right)}{2}+\class{steps-node}{\cssId{steps-node-3}{\dfrac{1}{2}}}\ln\left(|\tan\left(x\right)+\sec\left(x\right)|\right)\right)$$